Vocabulary
- Simple pendulum 单摆
- Amplitude 振幅 \(A\)
- Angular frequency 圆频率 \(\omega\)
- (Linear) frequency 频率 \(f\)
- Period 周期 \(T\)
- Phase 相位 \(\omega x + \varphi\)
- Initial phase 初始相位 \(\varphi\)
Math Time
The solution to the second-order differential equation
\[\frac {d^2 x} {dt^2} + b x = 0
\]
is
\[x(t) = C_1 \sin(\sqrt b x) + C_2 \cos(\sqrt b x)
\]
which is equivalent to (for non trivial solutions)
\[\boxed{x(t) = A \sin(\sqrt b x + \varphi)}
\]
SHM
https://en.wikipedia.org/wiki/Simple_harmonic_motion
Simple harmonic motion (SHM) is a special type of periodic motion an object experiences by means of a restoring force whose magnitude is directly proportional to the distance of the object from an equilibrium position and acts towards the equilibrium position.
Angular Frequency, Linear Frequency, and Period
The angular frequency of \(A \sin (\omega x + \varphi)\) is \(\omega\).
The linear frequency is defined as:
\[\boxed{f = \frac {\omega} {2 \pi}}
\]
The period, defined as the time per cycle, is the inverse of the frequency:
\[\boxed{T = \frac 1 f}
\]
\[\boxed{f = \frac {\omega} {2\pi} = \frac 1 T}
\]
Example 9.1 Simple Pendulum
Calculate the frequency of oscillation of a simple pendulum of length \(L\), assuming small oscillations.
(See Figure 9.4)
\[\begin{aligned}\tau = I \alpha &= - mgl \sin \theta \\I \alpha &\approx -mgl \theta & \text{for small oscillations} \\\frac {d^2 \theta} {dt^2} + \frac {mgl} I \theta &= 0 \\
\end{aligned}\]
Solving this second-order differential equation, we get:
\[\theta = A \sin\left( \sqrt{\frac {mgl} I} t + \varphi \right)
\]
Which means that the angular frequency is:
\[\omega = \sqrt {\frac {mgl} I} = \sqrt {\frac {mgl} {ml^2}} = \sqrt {\frac g l}
\]
And we can now calculate the linear frequency as well:
\[f = \frac \omega {2 \pi} = \frac 1 {2\pi} \sqrt {\frac g l}
\]
From this example, we derived the angular frequency for a simple pendulum (assuming small oscillations):
\[\boxed{\omega = \sqrt {\frac g l}}
\]
The \(k_{\text{effective}}\) method
We use the slope of \(F(x)\) or \(\tau(\theta)\) at \(x = 0\) or \(\theta = 0\) to approximate \(F(x)\) or \(\tau(\theta)\):
\[\boxed{k_{\text{effective}} = \left.-\frac {dF} {dx} \right|_{\text{equilibrium distance}}}
\]
\[\boxed{k_{\text{effective}} = \left.-\frac {d\tau} {d\theta} \right|_{\text{equilibrium distance}}}
\]
Then we can calculate the angular frequency for a SHM system:
\[\boxed{\omega = \sqrt {\frac {k_{\text{effective}}} m}}
\]
\[\boxed{\omega = \sqrt {\frac {k_{\text{effective}}} I}}
\]
Example 9.2 Mass-Spring System
What is the general mathematical description of a mass connected to a spring oscillating on a horizontal frictionless surface?
\[\begin{aligned}F = ma &= -kx \\\frac {d^2 x} {dt^2} + \frac k m x &= 0 \\
\end{aligned}\]
Solving this second-order differential equation, we get:
\[x(t) = A \sin\left( \sqrt {\frac k m} t + \varphi \right)
\]
Example 9.4
A \(2\) kg mass is oscillating on a spring with a spring constant of \(10\) N/m. When the spring is stretched \(1\) m, the velocity of the mass is \(13\) m/s. What is the amplitude of the oscillation?
Solution 1 (from the book)
Energy conservation (when the spring is maximally stretched, it has zero speed):
\[\begin{aligned}\frac 1 2 m v^2 + \frac 1 2 k x^2 &= \frac 1 2 k x_{\max}^2 \\\frac 1 2 (2kg) (13m/s)^2 + \frac 1 2 (10N/m) (1m)^2 &= \frac 1 2 (10N/m) x_{\max}^2 \\x_{\max} &= \sqrt{\frac {174} 5}m \approx \boxed{5.899m} \\
\end{aligned}\]
Solution 2
We know
\[x(t) = A \sin\left( \sqrt {\frac k m} t + \varphi \right)
\]
Differentiate both sides:
\[v(t) = A \cos\left( \sqrt {\frac k m} t + \varphi \right) \times \sqrt {\frac k m}
\]
Plug the given data into these formulas:
\[\begin{cases}x(0) = A \sin \varphi = 1m \\v(0) = A \cos \varphi \times \sqrt 5 s^{-1} = 13m/s \\
\end{cases}\]
\[A = \sqrt {\frac {174} 5}m \approx \boxed{5.899m}
\]
Example 9.5 Physical Pendulum
(See page 256)
(a)
From Example 9.1, we know \(\omega = \sqrt {\frac {mgl} I}\):
\[\omega = \sqrt {\frac {mgl} I} = \sqrt{40} s^{-1} \approx \boxed{6.325} s^{-1}
\]
(b)
\[U = mgh = mgd(1 - \cos \theta) = (20J) (1 - \cos\theta)
\]
(c) Solution 1 (from the book)
Energy conservation:
\[E_{\text{mech}} = U + KE = mgd(1 - \cos\theta) + \frac 1 2 I \omega^2
\]
\[(20J)(1 - \cos 0.1) + \frac 1 2 (0.5 kg \cdot m^2) (0.05 s^{-1})^2 = 0 + \frac 1 2 (0.5 kg \cdot m^2) \omega_{\max}^2
\]
\[\omega_{\max} = \frac {\sqrt {161}} {20} s^{-1} \approx \boxed{0.634 s^{-1}}
\]
(c) Solution 2
Follow the same way in Solution 2 of Example 9.4:
\[\begin{cases}\theta(1s) = A \sin \left( \sqrt{40} + \varphi \right) = 0.1 \\\tau(1s) = A \cos \left( \sqrt{40} + \varphi \right) \times \sqrt{40} s^{-1} = 0.05 s^{-1} \\
\end{cases}\]
\[A = \frac {\sqrt{1610}} {400}
\]
\[\omega_{\max} = A \times \sqrt{40} s^{-1} = \frac {\sqrt {161}} {20} s^{-1} \approx \boxed{0.634 s^{-1}}
\]
